A grain silo is shown below.
What is the volume of grain that could completely fill this silo rounded to the nearest whole number? Use 22/7 for pi.
19,008 ft3
19,461 ft3
6,336 ft3
453 ft3

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

##### 1 Attachment

- anonymous

Sum together half of a sphere for the top part, with a cylinder for the bottom part. That's all there is to it :-)
Do you know the formulas for a cylinder and a sphere?
h = 168 ft
r = 6 ft

- anonymous

no i don't

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

\[\large V_{sphere} = \frac{4 \pi r ^3}{3}\]
\[\large V_{cylinder} = \pi r^2 h\]
\[V_{total} = \frac{V_{sphere}}{2} + V_{cylinder}\]

- anonymous

Can you solve it now? ;-D

- anonymous

no i'm sorry i'm completely lost :-(

- anonymous

r = radius (both for the cylinder and sphere) , h = height of the cylinder
Separate the curved, hemispherical top from the cylinder under it. You'll just add volume the two shapes together. What are you confused on specifically?

- anonymous

i'm just naturally confused on everything that has to do with maths you have to talk to me like a toddler and tell mi step by step

- anonymous

Can you identify the radius in the drawing? And the height?

- anonymous

the radius for the entire drawing?

- anonymous

There's only one radius, yes. It works for both the sphere and cylinder, like I said.

- anonymous

is it 6 ft?

- anonymous

Yes...
|dw:1342123496209:dw|

- anonymous

And height is "h", which is?

- anonymous

168 ft

- anonymous

2/3*22/7*6*6*6 + 1/3*@22/7*6*6*168 = ............

- anonymous

Ok, now scroll up and put r & h into those two equations for the Volume of a Sphere and Cylinder. Substitute, replacing r with "(6 ft)", and replacing h with "(168 ft)".

- anonymous

ok

- anonymous

|dw:1342123834485:dw|
find the volume of given figure

- anonymous

@abhishekjha29 no, it's a cylinder under a hemisphere, not a cone.

- anonymous

ya tats ok.but i have a new figure wid a cone instead of culinder

- anonymous

so i use 3/4 3.14 times 6 to the 3rd power to find the volume of the sphere?

- anonymous

???
@abhishekjha29 the volume of a cone = \(\large\frac{\pi r ^2 h}{3}\) , exactly a third of a cylinder's volume with the same dimensions for height and radius.

- anonymous

oh ok

- anonymous

when i do it it tells me that the volume for the cylinder is 6330.24

- anonymous

@rudyjanay I would leave the 3.1415... \(\pi\) out of it until the end, just keep it along like it's a unit like "feet". You can indeed do that above, but keep in mind it will be an estimate because:
\(\pi \approx 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 062...\)

- anonymous

Best to just leave it alone until you're all done with the formulas :-D

- anonymous

Kind of like how I did in the cylinder part above, if you see what I did

- anonymous

When you're all done, multiply that big, non-decimal number and it'll give you your messy decimal answer.

- anonymous

ok i'll try leaving the 3.14 out

- anonymous

:-)
I think you've got it down now
It's just plug & chug once you got the setup up done. The tricky part was recognizing it's half of a sphere (a hemisphere) plus a cylinder added together, and that the cylinder and the hemisphere have the same radius.

- anonymous

oh i guess my phone was giving me the wrong answer because i don't have a calculator with me

- anonymous

the answer that i got is 19 461

- anonymous

I just noticed I made a typo let me correct myself...
\[\large V_{cylinder} = \pi r^2 h\]
\[\large V_{cylinder} = \pi (6 \ ft)^2 (168 \ ft)\]
\[\large V_{cylinder} = \pi (36 \ ft^2)(168 \ ft)\]
\[\large V_{cylinder} = 6048\pi \ ft^3\]
And back into the total...
\[\large V_{total} = \frac{288\pi \ ft^3}{2} + (6048\pi ft^3)\]
\[\large V_{total} = 144\pi \ ft^3 + 6048\pi ft^3\]
\[\large V_{total} = 6192\pi \ ft^3 \approx \ 19452.74171... ft^3\]

- anonymous

oh so i was right

- anonymous

So yes, you're good. The answer choice you were given is actually an estimate, and not as exact as what I just did :-)

- anonymous

ohhhh great thank you

- anonymous

6292\(\pi\) ft\(^3\) is the exact answer, and what I would be expect to do at work

- anonymous

Because if a construction firm wants to make an estimate to their liking, they can just pull out their pocket calculator or smartphone and see how many decimal places they care about for the project

- anonymous

Sometimes accuracy is critical and a matter of life & death, other times you don't have to be so particular

- anonymous

thats why i don't get maths

- anonymous

Example of where accuracy is a matter of life & death: ammunition factory.
Even the smallest screw up could mean a detonation, a dud bullet in a critical field operation that means some cop or soldier gets killed, contracts being canceled etc.
Another good example would be a factory where they make care engines. If they get the radius slightly wrong your engine won't work well or is more prone to breakdowns.
Math can seem useless until you have something that makes it mean something :-)

- anonymous

Car engines* lol not "care"

- anonymous

are you a math wiz or something

- anonymous

Engineering student. I'm not mathematician, but I love the application of it to stuff that matters :-3 Fire & explosives for example is a specialty of mine, and game design is my hobby (which also uses math quiet a bit because of physics stuff).

- anonymous

Math itself though, meh... I loose focus if it doesn't mean anything practical.

- anonymous

oh ok i'm gonna check on my other question

Looking for something else?

Not the answer you are looking for? Search for more explanations.